It was discrete uniform so probailities all 1/4. Three solutions. (1/4 x 1/4 x 1/4) = 1/64. Possible scores were 1 2 3 4. From there there's 4 unique combinations, 123, 234, 431, 134 (possibly more). So 1/64 x 4 = 1/16.

I think anyway.

Reply 6

Ladddd

2

@ProlongRS
I think that’s right but I’m pretty sure there are other like 2,3,1. 2,1,3. 3,2,4. I think there were like 16 probabilities so it turned out to be 1/64 times 18 which is an 1/8 I think

Reply 7

ProlongRS

OP

3

I think it was only 2 marks so I didn't spend much time on it

what did you guys get for 5 year regression estimation

Reply 8

ergsegsergseges

6

For the dice question you have 1/4 chance to roll any number, then it’s 3/4 to not roll it again and 3/4 to not roll your previous number, so you multiply those together
And then X3 as this can happen 3 times as you only chose 1/4 to start with (1 value)

Reply 9

Ladddd

2

Did it one way and got 270 (crossed it out) and then I did it another way I got 277.1 words which one is right?

Reply 10

Tarabanana123

2

Man I flopped this one so bad, didn’t even manage the histogram one :frown:

I’m sure I have a U so I can forget about maths now. I need an overal grade of a C

Reply 11

kparmar1999

6

What did u lot get for the normal distribution one I got 0.2033 n then the 2days bit I got 0.7116
The the new sd I got 4.27 something like this

Reply 12

xxxtentacion..

16

i got this tooo
i said another way of graphing is coding for the last q, and for the histogram using a normal distribution bc the skewness was almost 0
and the probability (last part) for the M value graph given x1 < 2 ans M=4, i got 2/7?

Original post by kparmar1999

What did u lot get for the normal distribution one I got 0.2033 n then the 2days bit I got 0.7116
The the new sd I got 4.27 something like this

Reply 13

xxxtentacion..

16

i kinda enjoyed that paper but looking on here people got diff answers to me :frown:

Reply 14

xxxtentacion..

16

great i didnt get that

Original post by ergsegsergseges

For the dice question you have 1/4 chance to roll any number, then it’s 3/4 to not roll it again and 3/4 to not roll your previous number, so you multiply those together
And then X3 as this can happen 3 times as you only chose 1/4 to start with (1 value)

Reply 15

thrizzy

8

New SD estimate was 4.82 not 4.27

Reply 16

xxxtentacion..

16

wb probability question?

Reply 17

thrizzy

8

To finish the venn diagram, you just used probability formula for p(BUC) and filled it in I think p(BandC) was 0.17.

Original post by xxxtentacion..

wb probability question?

Reply 18

orangestapler

10

Original post by ergsegsergseges

And for the love of god, the last last question, HOW WAS SHE SUPPOSED TO IMPROVE THE GRAPH.
And the other one for improving the histogram

I lit just had to put gain more data, had no clue. What did you get for the regression equation?

Reply 19

username2741103

8

How I answered: But then the third couldn't be the same as the 1st or 2nd. So I did 1/4 * 3/4 * 1/2 . As the 1/2 would be the 3rd could be 1 of the 2 numbers that wasn't picked originally. I then multiplied by 4 as could start with 1, 2, 3, or 4, but my fraction seemed to high for the probability of this to happen. (I think I'm wrong though, someone tell me the answer :biggrin:

!)

Or did I misunderstand. I read it as all 3 are different numbers.
(Original post by ergsegsergseges)

For the dice question you have 1/4 chance to roll any number, then it’s 3/4 to not roll it again and 3/4 to not roll your previous number, so you multiply those together
And then X3 as this can happen 3 times as you only chose 1/4 to start with (1 value)

Edexcel Maths S1 Resit Post-Exam

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